- 2075 Baisakh · 4 marks
Thevenin: Vth, Rth and 5 kΩ current
Short answer
Thevenin's theorem: Vth = 48×4/(12+4) = 12 V, Rth = 8k + (12k∥4k) = 11 kΩ. Current through 5kΩ is I = 12/(11k+5k) = 0.75 mA.
Determine Vth, Rth and the current through the 5 kΩ resistor using Thevenin's theorem. [Figure: 48 V source; 12 kΩ and 8 kΩ resistors in series along the top; a 4 kΩ resistor and the 5 kΩ resistor in vertical branches.]
Assumed circuit: the 48 V source feeds a 12 kΩ resistor in series along the top, reaching node A. A 4 kΩ resistor connects A to ground (vertical branch). An 8 kΩ resistor in series along the top connects A to node B, and the 5 kΩ load is connected from B to ground.
Step 1: remove the 5 kΩ and find Vth
With the 5 kΩ open, no current flows in the 8 kΩ, so Vth is the voltage across the 4 kΩ:
Vth = 48 × 4/(12 + 4) = 12 V
Step 2: find Rth
Short the 48 V source. Looking into terminals B-ground:
Rth = 8 kΩ + (12 kΩ ∥ 4 kΩ) = 8 + (12 × 4)/16 = 8 + 3 = 11 kΩ
Step 3: current through 5 kΩ
I = Vth/(Rth + RL) = 12 V/(11 kΩ + 5 kΩ) = 12/16 mA
Answer: Vth = 12 V, Rth = 11 kΩ, I(5 kΩ) = 0.75 mA
✓ Answer checked by Claude Opus 5.5 (high effort)
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